3[tex]\sqrt[6]{x+1} =\sqrt[3]{x+1}+2[/tex]

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To solve this equation, we first need to isolate the term containing the cube root on one side of the equation.

Let's raise both sides of the equation to the 6th power to eliminate the radicals:

tex^6 = (\sqrt[3]{x+1} + 2)^6[/tex]

This simplifies to:

[tex]3^6 (x+1) = (\sqrt[3]{x+1} + 2)^6[/tex]

Now we will substitute: let [tex]y = \sqrt[3]{x+1}[/tex]. Then:

[tex] (3y)^6 = (y + 2)^6[/tex]
[tex] 3^6y^6 = (y + 2)^6[/tex]
[tex] 729y^6 = (y + 2)^6[/tex]

Expanding both sides using binomial theorem:

[tex] 729y^6 = y^6 + 6y^5(2) + 15y^4(2)^2 + 20y^3(2)^3 + 15y^2(2)^4 + 6y(2)^5 + 2^6[/tex]

This simplifies to:

[tex] 729y^6 = y^6 + 12y^5 + 60y^4 + 160y^3 + 240y^2 + 192y + 64[/tex]

Moving all terms to one side, and combining like terms will result in a polynomial equation. This can then be solved for y, and then back-substituted to find the values of x that satisfy the original equation.